Prove that $F$ is differentiable on $\mathbb{R}$. Let $f$ be an  integrable function on $\mathbb{R}$, i.e.,  $ \displaystyle\int_{\mathbb{R}} |f| \ dx <\infty $.
And let $F(u)=\displaystyle \int_{\mathbb{R}} f(x) \sin(ux)\ dx$ where $u \in \mathbb{R}$ and then, $F$ is continuous on $\mathbb{R}.$
Suppose $\displaystyle\int_{\mathbb{R}} (1+|x|)|f(x)| dx < \infty.$
Then, prove that $F$ is differentiable on $\mathbb{R}$.
My attempt is here.
I'll prove $\displaystyle\lim_{h\to 0} \dfrac{F(a+h)-F(a)}{h}$ exists for all $a \in \mathbb{R}$.
$F(a+h)-F(a)=\displaystyle\int f(x)(\sin(a+h)x-\sin(ax)) dx$
Let $k_n(x)=n f(x) \left( \sin\left(a+\dfrac{1}{n}\right)x - \sin(ax) \right)$.
Then, $|k_n (x)|\leqq |f(x)|\big(1+|x| \big)$ and $\displaystyle\lim_{n \to \infty} k_n (x) = xf(x)\cos(ax)$ hold.
Thus, from Dominated convergence theorem, $\displaystyle\lim_{n \to \infty} \int k_n (x) dx =\int xf(x)\cos(ax) dx$.
Then,
\begin{align}
&\quad \ \displaystyle\lim_{h\to 0} \dfrac{F(a+h)-F(a)}{h}\\ &=\lim_{h\to 0} \int f(x) \dfrac{\sin(a+h)x-\sin(ax) }{h} dx \\
&=_{\text{Let } h=\frac{1}{n}} \lim_{n\to \infty}  \int n f(x) \left(\sin\left(a+\dfrac{1}{n}\right)x-\sin(ax)\right)  dx \\
&=\lim_{n\to \infty}  \int k_n (x) dx \\
&=\int xf(x)\cos(ax) dx.
\end{align}
So, if I can say $\int xf(x)\cos(ax) dx$ exists, I can end the proof, but I don't know how I can say it.
And I'm not sure my attempt is correct. I let $h=\dfrac{1}{n}$ in the limit calculation but is this possible ?
Thank you for your help.
 A: For $a, x \in \Bbb R$, we have $$|xf(x) \cos(ax)|\le |x f(x)| \le (1 + |x|)|f(x)|.$$
Thus, your hypothesis tells you that $$\int_{\Bbb R} |x f(x) \cos(ax)| \ {\mathrm d}x < \infty.$$
In turn, the integral you were contemplating about, also exists.

However, the other problem you mention does exist. What you have shown is only that the limit $$\lim_{n \to \infty} \frac{F\left(a + \frac{1}{n}\right) - F(a)}{\frac{1}{n}}$$
exists for all $a \in \Bbb R$.
The above is not enough, though. (For example, consider $F$ to be the indicator function of $\Bbb Q$. Then, the above limit exists and is $0$ for all $a \in \Bbb R$ but $F$ is not even continuous anywhere.)
However, we can do the following: Fix $a \in \Bbb R$.
To show that
$$\lim_{h \to 0} \frac{F(a + h) - F(a)}{h}$$
exists, it suffices to prove that
$$\lim_{n \to \infty} \frac{F(a + h_n) - F(a)}{h_n}$$
exists for every sequence $(h_n)_n$ converging to $0$.

Now, going back to your situation, simply modify your proof by considering any arbitrary sequence $(h_n)_n$ converging to $0$ and replace $\frac{1}{n}$ with $h_n$ everywhere and you'll be done.
In fact, this is what one usually does when dealing with real limits because to use DCT, one has to go via sequences.
